Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{x - 4}{-10x + 20} \times \dfrac{6x - 54}{x^2 - 13x + 36} $
Solution: First factor the quadratic. $q = \dfrac{x - 4}{-10x + 20} \times \dfrac{6x - 54}{(x - 4)(x - 9)} $ Then factor out any other terms. $q = \dfrac{x - 4}{-10(x - 2)} \times \dfrac{6(x - 9)}{(x - 4)(x - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (x - 4) \times 6(x - 9) } { -10(x - 2) \times (x - 4)(x - 9) } $ $q = \dfrac{ 6(x - 4)(x - 9)}{ -10(x - 2)(x - 4)(x - 9)} $ Notice that $(x - 9)$ and $(x - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 6\cancel{(x - 4)}(x - 9)}{ -10(x - 2)\cancel{(x - 4)}(x - 9)} $ We are dividing by $x - 4$ , so $x - 4 \neq 0$ Therefore, $x \neq 4$ $q = \dfrac{ 6\cancel{(x - 4)}\cancel{(x - 9)}}{ -10(x - 2)\cancel{(x - 4)}\cancel{(x - 9)}} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $q = \dfrac{6}{-10(x - 2)} $ $q = \dfrac{-3}{5(x - 2)} ; \space x \neq 4 ; \space x \neq 9 $